Question

Take the mean distance of the moon and the sun from the earth to be 0.4 106 km and 150 106 km respectively. Their masses are 8 1022 kg and 2 1030 kg respectively. The radius of the earth is 6400 km. Let f1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and f2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to is

Answer 1

Answer:

Take the mean distance of the moon and the sun from the earth to be 0.4 106 km and 150 106 km respectively. Their masses are 8 1022 kg and 2 1030 kg respectively. The radius of the earth is 6400 km. Let f1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and f2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to is

Answer 2

Explanation:

Given Take the mean distance of the moon and the sun from the earth to be 0.4 106 km and 150 106 km respectively. Their masses are 8 1022 kg and 2 1030 kg respectively. The radius of the earth is 6400 km. Let f1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and f2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to is

• We need to find the ratio of f1 / f2
• Now force f1 = - G m me / r1^2
• Similarly f2 = G me ms / r2^2
• Now Δf1 = -Gmme d(r1^-2)
• So Δf1 = -Gmme / r1^3 Δr1
• Similarly Δf2 = - 2 Gme ms / r2^3 Δr2
• So now we need to find the ratio.
• So Δf1 / Δf2 = mΔr1 / r1^3 x r2^3 / ms Δr2
• Also Δr1 = Δr2 = 2
• Now substituting the values we get
• 8 x 10^22 x 2 x (150)^3 x 10^18 / (0.4 x 10^6)^3 x 2 x 10^30 x 2
• 54000000 / 0.256 x 10^8
• We get 2.10

Reference link will be

https://brainly.in/question/11372134